$$E=k\frac{Q}{r^2}$$

Where:

$E$ is the electric field strength in Newtons per coulomb.

$k$ is the Coulomb constant equal to $8.988\times10^9$

$Q$ is magnitude of the charge in coulomb.

$r$ is the distance from the charge in meters.

The Earth has a charge of $-6.0\times10^{13} C$ with its radius \(6.378\times10^6\)m

Substituting to calculate electric field \(E\):

$$E=8.988\times10^9\frac{6.0\times10^{13}}{(6.378\times10^6)^2}$$

$$E= 1.34\times10^2 $$$$V/m$$